/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 The Taipei 101 Tower in Taipei, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 80

The Taipei 101 Tower in Taipei, Taiwan is a \(509-\mathrm{m}\) \((1667-\mathrm{ft}), 101\) -story building ( \(\mathbf{F i g} .2 .28)\). The outdoor observation deck is on the 89 th floor, and two high- speed elevators that service it reach a peak speed of \(1008 \mathrm{~m} / \mathrm{min}\) on the way up and \(610 \mathrm{~m} / \mathrm{min}\) on the way down. Assuming these peak speeds are reached at the midpoint of the run and that the accelerations are constant for each leg of the runs, (a) what are the accelerations for the up and down runs? (b) How much longer is the trip down than the trip up?

Short Answer

Expert verified
The acceleration for upwards is approximately 2000 m/min虏, and for downwards about 732 m/min虏. The downward journey takes about 1.5 minutes longer.

Step by step solution

01

Determine the Total Distance of the Elevator Journey

The elevator journey is from the ground floor to the 89th floor of the Taipei 101 Tower. The building itself is 509 meters tall, but since we are traveling only to the 89th floor, let's assume that it covers approximately the same linear distance for clarity since this height was not specifically provided.
02

Divide the Journey into Two Equal Halves

Since the peak speeds are reached at the midpoint, the journey can be divided into two parts: acceleration to the midpoint and deceleration after the midpoint. The midpoint, therefore, is at half of the total distance traveled.
03

Calculate Time Using Uniform Acceleration Equations

Use the kinematic equation: \[v^2 = u^2 + 2as\]Where:- \(v = 1008\, \text{m/min}\) (up) and \(610\, \text{m/min}\) (down) are the final velocities,- \(u = 0\, \text{m/min}\) (initial velocity),- \(s = \frac{509}{2} = 254.5\, \text{m}\) (half the journey since the midpoint),- \(a\) is acceleration.Use this formula to compute acceleration for both upwards and downwards runs.
04

Solve for Upward Acceleration

Plug into the kinematic equation for the upward journey:\[(1008)^2 = 0 + 2a(254.5)\]Solve for \(a\):\[1008^2 = 2a(254.5)\] \[a = \frac{1008^2}{2\times 254.5}\]Compute to find the acceleration upwards.
05

Solve for Downward Acceleration

Use the same kinematic equation for the downward journey:\[(610)^2 = 0 + 2a(254.5)\]Solve for \(a\):\[610^2 = 2a(254.5)\] \[a = \frac{610^2}{2\times 254.5}\]Compute to find the acceleration downwards.
06

Determine Total Time for Up and Down Journeys

Use the formula \(v = u + at\) to solve for the time taken to reach peak speed:\[t = \frac{v - u}{a}\]Calculate \(t\) for both up and down journeys using their respective accelerations from Steps 4 and 5, considering accelerating and decelerating parts separately and summing them for total time.
07

Calculate the Difference in Time Between Journeys

Find the difference in total time by subtracting the upward journey time from the downward journey time calculated in Step 6.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Uniform acceleration occurs when the velocity of an object changes at a constant rate. This means the object either speeds up or slows down steadily over time. In our elevator problem, as the elevator travels to its peak speed at the midpoint going up and down, it does so with uniform acceleration. This concept is fundamental because it allows us to predict motion using straightforward equations.
  • Acceleration is constant: this simplifies computations and predictions of motion.
  • It involves two segments: going from rest to the peak speed, and back to rest.
  • Uniform acceleration is described mathematically using kinematic equations.
Understanding uniform acceleration helps in solving real-world motion problems by breaking them into predictable segments of motion.
Kinematic Equations
Kinematic equations are essential tools for solving motion problems involving uniform acceleration. They relate different physical quantities like velocity, acceleration, displacement, and time. In the context of our elevator example, the key kinematic equation \[v^2 = u^2 + 2as\]is used. Here:
  • \(v\) is the final velocity that the elevator reaches (1008 m/min going up, 610 m/min coming down).
  • \(u\) is the initial velocity, initially 0 since the elevator starts from rest.
  • \(a\) is the acceleration we want to find, different for the upward and downward journeys.
  • \(s\) is the displacement to the midpoint.
Through this equation, we compute the acceleration values for ascending and descending journeys. Other equations such as \(v = u + at\) are used to find the time taken, linking acceleration to time.
Elevator Physics
Elevator physics involves understanding how elevators, through gravitational potential energy and external forces, are efficiently moved between floors in tall buildings. Elevators reach high speeds for convenience, and their motion needs to be thoroughly managed through controlled accelerations. In the Taipei 101 example:
  • The elevator reaches its peak speed at 1008 m/min upwards and 610 m/min downwards.
  • Acceleration and deceleration need to be applied smoothly to ensure safety and comfort.
  • The trip is analyzed in segments to find the acceleration and time differences.
By considering these physical principles, engineering efficient and safe elevators becomes practical. Elevator physics incorporates mechanics, safety, and engineering design considerations into creating high-speed transits.
Problem Solving in Physics
Problem solving involves understanding the scenario, breaking down the problem, and methodically applying physics principles to find solutions. With our elevator problem, follow these steps:
  • Break down the trip into acceleration and deceleration phases since peak speed is reached midway.
  • Use kinematic equations: identify what information is given and what needs to be found.
  • Calculate in pieces: first find acceleration during each trip phase, then time, finally compare upwards and downward journeys.
Problem solving in physics is systematic; it requires careful reading of the problem, a good grasp of relevant formulas, calculations, and interpreting the results. The process not only solves the given problem but also strengthens the conceptual understanding.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An automobile traveling at \(15.0 \mathrm{~km} / \mathrm{h}\) along a straight, level road accelerates to \(65.0 \mathrm{~km} / \mathrm{h}\) in \(6.00 \mathrm{~s}\). What is the magnitude of the auto's average acceleration?

A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car slams on her brakes and skids on dry pavement for \(50 \mathrm{~m},\) decelerating at \(7.5 \mathrm{~m} / \mathrm{s}^{2}\). Then she hits the icy patch and skids another \(80 \mathrm{~m}\) before coming to rest. If her initial speed was \(70 \mathrm{mi} / \mathrm{h}\), what was the deceleration on the ice?

An object initially at rest experiences an acceleration of \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{~s}\) and then travels at that constant velocity for another \(8.0 \mathrm{~s}\). What is the object's average velocity over the 14 -s interval?

A roller coaster car traveling at a constant speed of \(20.0 \mathrm{~m} / \mathrm{s}\) on a level track comes to a straight incline with a constant slope. While going up the incline, the car has a constant acceleration of \(0.750 \mathrm{~m} / \mathrm{s}^{2}\) in magnitude. (a) What is the speed of the car at \(10.0 \mathrm{~s}\) on the incline? (b) How far has the car traveled up the incline at this time?

In an air bag test, a car traveling at \(100 \mathrm{~km} / \mathrm{h}\) is remotely driven into a brick wall. Suppose an identical car is dropped onto a hard surface. From what height would the car have to be dropped to have the same impact as that with the brick wall?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Kinematics Physics

Read Explanation

Scientific Method Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.